Problem: There are values $A$ and $B$ such that
\[\frac{Bx-11}{x^2-7x+10}=\frac{A}{x-2}+\frac{3}{x-5}.\]Find $A+B$.
Explanation: We factor the denominator in the left-hand side to get \[\frac{Bx - 11}{(x - 2)(x - 5)}= \frac{A}{x - 2} + \frac{3}{x - 5}.\] We then multiply both sides by $(x - 2)(x - 5)$, to get \[Bx - 11 = A(x - 5) + 3(x - 2).\] We can solve for $B$ substituting a suitable value of $x$.  For example, setting $x = 5$, the equation becomes $5B - 11 = 9$, so $B = 4$.  Then \[4x - 11 = A(x - 5) + 3(x - 2).\] Setting $x = 2$, this equation becomes $-3 = -3A$, so $A = 1$.  Therefore, $A + B = 1 + 4 = \boxed{5}$.